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import Foundation
/*:
 编辑距离
 给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数 。

 你可以对一个单词进行如下三种操作：

 插入一个字符
 删除一个字符
 替换一个字符

 输入：word1 = "horse", word2 = "ros"
 输出：3
 解释：
 horse -> rorse (将 'h' 替换为 'r')
 rorse -> rose (删除 'r')
 rose -> ros (删除 'e')

 输入：word1 = "intention", word2 = "execution"
 输出：5
 解释：
 intention -> inention (删除 't')
 inention -> enention (将 'i' 替换为 'e')
 enention -> exention (将 'n' 替换为 'x')
 exention -> exection (将 'n' 替换为 'c')
 exection -> execution (插入 'u')
 */


/*:
 动态规划
 if s1[i] == s2[j]:
    skip
    i,j移动
 else:
    插入
    删除
    替换
 */
func minDistance(_ word1: String, _ word2: String) -> Int {
    let s1 = word1.map{ "\($0)" }
    let s2 = word2.map{ "\($0)" }
    let m = s1.count, n = s2.count
    var dp: [[Int]] = Array(repeating: Array(repeating: 0, count: n+1), count: m+1)
    // base case
    for i in 1...m {
        dp[i][0] = i
    }
    for j in 1...n {
        dp[0][j] = j
    }
    // 自底向上求解
    for i in 1...m {
        for j in 1...n {
            if s1[i-1] == s2[j-1] {
                dp[i][j] = dp[i-1][j-1]
            } else {
                dp[i][j] = min(dp[i-1][j]+1, min(dp[i][j-1]+1, dp[i-1][j-1]+1))
            }
        }
    }
    return dp[m][n]
}
print(minDistance("horse", "ros"))

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